Question
Establish the relation between critical angle and refractive index.
✓
Answer
Relation between Refractive Index and Critical Angle : Suppose rarer medium is denoted by 1 and denser medium by 2. For refraction of light from denser medium to rarer medium the refractive index of rarer medium w.r.t. denser medium is given by :
$_{2}n_{1}=\frac{\sin i}{\sin r}$ [By Snell's law]
where i = Angle of incidence in denser medium, and r = Angle of refraction in rarer medium.
But according to definition of critical angle: for
$\angle r=90^{\circ}$ we have
$i=i_{c}$
(where $i _{ c }=$ critical angle at interface of two media)
$\therefore \quad{ }_2 n_1=\frac{\sin i_c}{\sin 90^{\circ}} \Rightarrow{ }_2 n_1=\frac{\sin i_c}{1}$
$\Rightarrow \quad{ }_2 n_1=\sin i_c$ ... (1)
But according to principle of reversibility of light, we have
${ }_1 n_2=\frac{1}{{ }_2 n_1}$
$\Rightarrow \quad{ }_1 n_2=\frac{1}{\sin i_c}$ ...(2)
i.e. refractive index of denser medium with respect to rarer medium is equal to the reciprocal of the sine of critical angle at interface of two media.
$_{2}n_{1}=\frac{\sin i}{\sin r}$ [By Snell's law]
where i = Angle of incidence in denser medium, and r = Angle of refraction in rarer medium.
But according to definition of critical angle: for
$\angle r=90^{\circ}$ we have
$i=i_{c}$
(where $i _{ c }=$ critical angle at interface of two media)
$\therefore \quad{ }_2 n_1=\frac{\sin i_c}{\sin 90^{\circ}} \Rightarrow{ }_2 n_1=\frac{\sin i_c}{1}$
$\Rightarrow \quad{ }_2 n_1=\sin i_c$ ... (1)
But according to principle of reversibility of light, we have
${ }_1 n_2=\frac{1}{{ }_2 n_1}$
$\Rightarrow \quad{ }_1 n_2=\frac{1}{\sin i_c}$ ...(2)
i.e. refractive index of denser medium with respect to rarer medium is equal to the reciprocal of the sine of critical angle at interface of two media.
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